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For a real function $g$ on $X\times Y\times Z$ and functions $a(y,z),b(x,z),c(x,y)$ bounded in absolute value by one,First apply Cauchy-Schwarz inequality in $(y,z)$ to remove $a$, duplicating $x$. Apply it in $(x_0,x_1,z)$ to remove $b(x_0,z)b(x_1,z)$, duplicating $y$. Finally apply it in $(x_0,x_1,y_0,y_1)$ to remove the four $c$ factors, duplicating $z$. The resulting eighth-power bound is exactly the eight-factor average defining the three-dimensional box norm. Functions depending on just one coordinate, or constants, can be incorporated in any of the three pair factors.

A pairing function is a computable function that gives a bijection from the Cartesian product ${\mathbb{N}}_0^2$ to ${\mathbb{N}}_0$, with two computable functions recovering its coordinates. Here ${\mathbb{N}}_0=\{0,1,2,\dots \}$ consists of the natural numbers, including zero. If $p$ is the pairing function and $L,R$ its decoding functions, the requirements areEvery ordered pair has exactly one code, every code represents an ordered pair, and decoding always terminates. For a bijection given by a total computable function, computability of the inverse function also follows by enumerating all ordered pairs until the one with code $z$ appears. A primitive recursive pairing function makes the stronger requirement that both encoding and coordinate decoding are primitive recursive.

The Cantor pairing function makes this concrete through triangular numbers:On the diagonal $x+y=s$, list $(s,0),(s-1,1),\dots ,(0,s)$. There are $s+1$ ordered pairs there, and $T(s)$ ordered pairs on earlier diagonals. Their codes therefore occupy precisely the consecutive natural numbers from $T(s)$ through $T(s+1)-1$. These intervals touch without overlapping and cover all natural numbers; within each interval the offset $y$ determines the ordered pair. This proves both injectivity and surjectivity, hence the bijection. It also explains why the infinite grid is a countably infinite set: every ordered pair is reached after finitely many steps. A row-by-row listing would never finish its first infinite row. Authored treatments give the same diagonal construction with the coordinates exchanged.^[ref]

For example, the Cantor pairing function givesThe order of the coordinates matters: $(1,0)$ and $(0,1)$ are different ordered pairs with different codes.

To decode a natural number $z$, find the unique diagonal index $s$ satisfyingThe inequality gives $0\le y\le s$, so both recovered coordinates are natural numbers. At $z=18$, $T(5)=15\le 18<T(6)=21$, hence $s=5$, $y=18-15=3$, and $x=5-3=2$. At $z=0$, $s=0$ and both coordinates are zero. Solving the quadratic inequality gives the equivalent floor function expressionThis is exact as a statement about the real numbers and their square roots. An implementation using floating-point arithmetic can round across a diagonal boundary, especially for large $z$. Use an exact integer square root instead: with $q=\lfloor \sqrt{8z+1}\rfloor$ computed using integer arithmetic, $s=\lfloor (q-1)/2\rfloor$. Alternatively, search for the diagonal using exact integer comparisons. The primitive recursive pairing function proof below uses bounded minimization and needs no approximate square root.

A pairing function is a reversible encoding: unlike a hash function mapping arbitrarily many inputs into a fixed finite range, it loses no coordinates. This works because the mathematical natural numbers have no upper bound. A machine word with $b$ binary digits has $2^b$ possible values, while two such words have $2^{2b}$ possible ordered pairs; a bijection into one word of the same width is impossible. Evaluating the Cantor pairing function in a fixed-width integer type can overflow, including in the intermediate product $(x+y)(x+y+1)$. Exact implementation requires sufficiently large integer storage or an explicit restricted input range. One mathematical code need not occupy fewer binary digits than its two inputs.

Iterating a pairing function gives a reversible encoding of tuples of any fixed positive length. With the left-nesting convention, the Cantor pairing function encodes $(2,3,4)$ asDecode the outer ordered pair first: $T(22)\le 257<T(23)=276$ gives $(18,4)$. Decode its first coordinate $18$ to recover $(2,3)$, giving the nested ordered pair $((2,3),4)$ and thus the original tuple $(2,3,4)$.

The tuple length and nesting convention must be known. The same code $257$ represents $(18,4)$ as a two-entry tuple and $(2,3,4)$ as a three-entry tuple; nesting alone does not make variable-length tuples self-describing. To encode variable-length tuples, supply their length or another explicit termination convention as part of the encoding. This simple change of representation is useful in computability theory and Gödel numbering: a function of several natural numbers can be studied through a single numerical input, while all coordinates remain recoverable. The striking fact is that a whole infinite grid can be serialized without omitting a point.

A historical reference for comparing sets through bijections is Georg Cantor's paper Ein Beitrag zur Mannigfaltigkeitslehre, published in 1878 in volume 84 of Journal für die reine und angewandte Mathematik, pages 242–258.^[ref]

A parametric equation expresses coordinates as functions of one or more independent parameters. For example, $x=\cos t$, $y=\sin t$ describes a circle; multiple parameters describe surfaces. A smooth curve is a differentiable single-parameter realization. A reparametrization changes the parameter without changing the geometric image.

A partial derivative measures the local response of a function to one input while its other inputs stay fixed. For a function $f$ on an open subset of ${\mathbb{R}}^2$, its partial derivatives at $(a,b)$ are the limitswhen the respective limits exist. Each is an ordinary derivative of a one-variable slice: $x\mapsto f(x,b)$ or $y\mapsto f(a,y)$. This is why the other variable is treated as a constant during partial differentiation.^[ref]

For an original example, take the polynomial functionHolding $y$ fixed gives $f_x=2x+y$, while holding $x$ fixed gives $f_y=x+4y$. At $(1,1)$, the height is $f=4$ and the two partial derivatives are $f_x=3$ and $f_y=5$. The values answer different experiments on the same mathematical model: change $x$ alone, or change $y$ alone.

The two exact slices through that point areDividing either height change by $t\ne 0$ gives $3+t$ or $5+2t$; taking the limit $t\to 0$ recovers the two slopes. At $t=0.1$, the finite height changes are $0.31$ and $0.52$, although their linear approximations are $0.3$ and $0.5$. A derivative measures the limiting response, rather than an exact finite change.

For a function of $n$ variables, ${\partial }_{j}f(a)$ varies only its $j$th coordinate: it is the same difference quotient along $e_j$, the $j$th standard basis vector. This gives a practical sensitivity analysis: one can compare how a mathematical model responds to changes in its individual inputs, provided the chosen system of units and the quantities held fixed are stated. Comparing the numerical sizes of two partial derivatives in different units does not by itself identify a physically preferable input.

Existence of the partial derivatives at one point is weaker than differentiability. The example under partial derivatives do not imply continuity shows how both coordinate tests can miss a bad diagonal approach. When the first partial derivatives are continuous in a neighborhood, they do give differentiability there, allowing the total differential to combine all small input changes.

A partial differential equation relates an unknown function of several independent variables to its partial derivatives. For $u(x,t)$, the notation $u_t=\partial u/\partial t$ means changing time while holding position fixed; $u_{xx}={\partial }^{2}u/\partial x^2$ measures how its spatial slope changes. Unlike an ordinary differential equation, a partial differential equation can describe an entire evolving profile rather than a single trajectory.

The heat equation $u_t=D{u}_{xx}$ smooths spatial variations, the transport equation $u_t+c{u}_x=0$ moves them, and Laplace's equation $\Delta u=0$ describes a spatial equilibrium. Here $D>0$ is a diffusivity, $c$ is a velocity, and $\Delta$ is the Laplacian. The periodic heat and transport comparison makes the difference visible using identical initial conditions.

A partial differential equation usually supplies a local rule rather than a complete answer. For example, every $u(x,t)=A+Bx$ satisfies the heat equation with zero PDE source term; data select which profile is relevant. A complete problem specifies its domain of a function, coefficients, initial conditions, boundary conditions, and meaning of solution of a partial differential equation. A well-posed problem has existence, uniqueness, and continuous dependence on initial data in stated spaces; the equation alone guarantees none of these.^[ref]

Assume $a<b$. On continuous functions on this compact interval, both proposed distances are finite and symmetric, and vanish on identical functions. For the supremum norm, zero distance plainly implies pointwise equality. For the L2 norm, if a continuous difference is nonzero at any point, it stays bounded away from zero on a subinterval of positive length. Its squared integral is then positive, so zero distance also implies equality.

For the first metric, the pointwise triangle inequality followed by the supremum gives $d_1(x,z)\le d_1(x,y)+d_1(y,z)$. For the second, put $u=x-y$, $v=y-z$. The Cauchy-Schwarz inequality givesTaking square roots proves the required triangle inequality. Thus both are metrics.

The continuous-function space is complete in the uniform metric. If $(x_n)$ is a Cauchy sequence in $d_1$, at each $t$ its scalar values have a limit $x(t)$, by completeness of the scalar field. Given $\epsilon >0$, choose $N$ with $\Vert x_n-x_m{\Vert }_{\infty }<\epsilon$ for $m,n\ge N$. Letting $m\to \infty$ gives $|x_n(t)-x(t)|\le \epsilon$ uniformly in $t$. Hence there is uniform convergence. To see that the limit is continuous at $t_0$, approximate it uniformly by one continuous $x_n$ and useThe outer terms can be made small uniformly, and the middle term is small near $t_0$. This proves the needed uniform limit theorem and completeness.

For the second metric, let $H$ be zero on $(-1,0)$ and one on $(0,1)$; its value at zero is immaterial to the integral. The supplied ramp functions are continuous and satisfyAlthough $H$ is not in the continuous-function space, the integral triangle inequality yields $d_2(x_m,x_n)\le (3m{)}^{-1/2}+(3n{)}^{-1/2}\to 0$. Thus the sequence is Cauchy. If it converged in $d_2$ to a continuous $x$, the same inequality would give $\Vert x-H{\Vert }_2=0$. Continuity would then force $x=0$ on the negative half-interval and $x=1$ on the positive half-interval, contradicting continuity at zero. Therefore the continuous-function space is not complete in $d_2$; its limit in the larger L2 space is the discontinuous step.

Substituting $y=x^p$ in the homogeneous ordinary differential equation gives $p(p-1)-2=(p-2)(p+1)=0$. ThusFor a Green's function decaying at the two ends, use $u(x)=x^2$ on the side of zero and $v(x)=x^{-1}$ on the side of infinity. Their Wronskian is $u{v}^{\prime }-u^{\prime }v=-3$. The function must be continuous at $x=\xi$, or its second derivative would contain a derivative of a Dirac delta. Its derivative jump must be one, to give the specified positive delta source. These two conditions yield the boundary-decaying Green kernel for an inverse-square differential operator:Its derivatives at the join are $-2/3$ on the left and $1/3$ on the right, confirming the jump one. Away from the join each branch solves the homogeneous equation, and both endpoint conditions hold.

Integrate this Green's function against the forcing, whose support is between zero and one. For $0<x\le 1$,For $x\ge 1$, only the right branch contributes. ThereforeOn the first branch, $y^{\prime }=(2x/3)\log x+x/9$ and $y^{\prime \prime }=(2/3)\log x+7/9$, so $y^{\prime \prime }-2y/x^2=1$. The second branch is a multiple of the homogeneous solution $x^{-1}$, so its forcing is zero. At $x=1$, both $y$ and $y^{\prime }$ agree, with values $-1/9$ and $1/9$, so there is no additional delta source. Finally $x^2\log x\to 0$ as $x\downarrow 0$ and $x^{-1}\to 0$ as $x\to \infty$. A homogeneous solution obeying both boundary conditions has $C=D=0$, establishing uniqueness.

Not uniformly continuous. ChooseTheir separation is $\pi /(x_n+y_n)\to 0$, while $\sin (x_n^2)=1$ and $\sin (y_n^2)=-1$. Thus the function values stay distance two apart, contradicting uniform continuity. Boundedness of the function does not prevent increasingly rapid oscillations.

A function $f:I\to \mathbb{R}$ is uniformly continuous ifThe crucial point is that $\delta$ works throughout the interval, not just near one chosen point.

Yes, it is bounded on the closed unit interval. Choose $\delta$ for $\epsilon =1$, and choose finitely many grid points $t_j$ such that every $x\in [0,1]$ is within $\delta$ of some $t_j$. Then $|f(x)|\le 1+{\max }_j|f(t_j)|$, a finite bound. Equivalently, uniform continuity implies continuity, and a continuous real function on a compact set is bounded.

Yes, a bounded derivative gives uniform continuity on the half-line. If $|f^{\prime }|\le K$, the mean value theorem gives $|f(x)-f(y)|\le K|x-y|$. This is a Lipschitz condition, so for $K>0$ use $\delta =\epsilon /K$. If $K=0$, the function is constant. As usual for a derivative on this closed half-line, continuity at its endpoint is included.

For the real even function, let $s_N=a_0/2+{\sum }_{n=1}^N{a}_n\cos nx$. Fourier orthogonality gives ${\int }_{-\pi }^{\pi }\cos nxdx=0$ for $n\ge 1$ and ${\int }_{-\pi }^{\pi }\cos nx\cos mxdx=\pi {\delta }_{nm}$. ConsequentlyFor an even square-integrable function, the complete Fourier cosine series converges in the L2 norm; hence its norms converge. Taking $N\to \infty$ proves Parseval's identity in the required normalization:This limit argument avoids multiplying an infinite series without a justification of L2 norm convergence.

For the quadratic function, the Fourier coefficients areThe last equality follows from two integrations by parts, using $\sin n\pi =0$. ThusThis Fourier series is absolutely convergent; the periodically extended quadratic is continuous, so it represents its values also at the endpoints. Substituting its coefficients into Parseval's identity gives

Write the spline interpolation operator as $P_{\mathbf{x}}=T{A}_{\mathbf{x}}^{-1}R$, where $R$ samples at the interpolation sites and $T$ synthesizes the normalized B-spline expansion. Sampling has operator norm at most one, and nonnegative B-splines with sum at most one give $\Vert T\Vert \le 1$. HenceHere the matrix norm is the maximum absolute row sum. Existence of the inverse matrix requires the usual ordered distinct interpolation sites and the Schoenberg–Whitney theorem support conditions, with endpoint values interpreted by their one-sided limits. If the displayed diagonal condition is read without the implicit ordering and distinctness, it is insufficient: all three sites equal to $1/2$ in the quadratic example give positive diagonal entries but three identical rows. The estimate applies whenever the interpolating linear map in the question is defined uniquely.

For the quadratic Bernstein basis, $N_1=(1-x{)}^2$, $N_2=2x(1-x)$ and $N_3=x^2$. Sampling and inverting giveThe three cardinal Lagrange interpolation polynomials areThe Lebesgue constant of interpolation gives the exact operator norm: the upper bound follows from $|Pg(x)|\le \Vert g{\Vert }_{\infty }{\sum }_i|{\ell }_i(x)|$. To attain it at a maximizing point, prescribe at the three sites the signs of the corresponding cardinal polynomials and extend those values by a continuous piecewise linear function of norm one.

On $0\le x\le 1/2$, the first two cardinal polynomials are nonnegative and the last is nonpositive. Since their sum is one, their absolute sum is $1-2{\ell }_2(x)=1+2x-4x^2$. Its maximum is $5/4$ at $x=1/4$. Reflection gives the same maximum at $3/4$ on the other half. In particular the data $(1,1,-1)$ produce $1+2x-4x^2$, attaining $5/4$. ThusThe coefficient estimate loses cancellation among the B-splines, which explains why it is not sharp for the resulting function norm.

The Korovkin theorem says that a sequence of positive linear operators on continuous functions on $[0,1]$ converges uniformly to the identity on every continuous function if it does so on $1$, $x$ and $x^2$. Necessity is immediate; the quadratic barrier argument establishing sufficiency is given in Solution 6.

There is an endpoint misprint in the original PDF: the repeated knots at the right end must equal $1$, not $0$. Otherwise the advertised nondecreasing spline knot sequence on $[0,1]$ does not exist. Use the corrected clamped spline knot sequences, with fixed order $k$ and maximum gap $h_n\to 0$. Let $m_n$ denote the number of B-splines for the $n$th sequence. Nonnegativity and partition of unity show that the Schoenberg spline operator $V_n$ is positive and $V_{n}1=1$.

The monomial B-spline coefficients reproduce the first two nonconstant test polynomials. PutThe pair sum uses $r<s$, as in the PDF; the converted TeX incorrectly includes $r=s$. Every sampling point ${\tau }_j$ and every knot entering these coefficients lie in $[t_j,t_{j+k}]$, whose length is at most $k{h}_n$. Thus $|{\tau }_j-a_{1,j}|\le k{h}_n$. Also all these numbers belong to $[0,1]$, soThe positive B-spline weights giveTogether with $V_{n}1=1$, the three limits proveUnder the usual interior knot multiplicities at most $k-1$, the splines are continuous and the quoted form of the Korovkin theorem applies directly. If full interior multiplicity is allowed, the output can be discontinuous. The same positive-operator quadratic barrier proof still works with bounded output functions and the supremum norm, and proves exactly the stated uniform error limit without asserting continuity of each output. Indeed the stronger local-support error bound for a Schoenberg spline operator, $\Vert V_{n}g-g{\Vert }_{\infty }\le \omega (g,k{h}_n)$, follows by comparing $g(x)$ with each active sample value. The restriction $k\ge 3$ is needed for the displayed quadratic reproduction argument, not for this direct support estimate.